Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Organic Chemistry 2 > Carbohydrates (Chapter 24) > Flashcards

Carbohydrates (Chapter 24) Flashcards

You may prefer our related Brainscape-certified flashcards:
Periodic Table flashcards Chemistry 101 flashcards AP® Chemistry flashcards Organic Chemistry 101 flashcards
1
Q

Empirical Formula of Carbohydrates

A

Cm(H2O)n

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Monosaccharide

Simple Sugar

A

An aldehyde/ketone containing at least three carbon atoms and two hydroxyl groups.

A monosaccharide can be either an aldose OR a ketose.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Linkage Connecting Two Monosaccharides

A

Ether Linkage

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Aldose vs. Ketonse

A
  • Aldose: A monosaccharide containing an aldehyde group.
  • Ketose: A monosaccharide containing a ketone group.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Nomenclature: Monosaccharide Chain Length

A
  • Triose: 3 Carbons
  • Tetraose: 4 Carbons
  • Pentose: 5 Carbons
  • Hexose: 6 Carbons
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Numbering: Aldose vs. Ketose

A
  • Aldose: Numbering begins at the aldehydic carbon.
  • Ketose: Numbering begins at the terminal carbon closest to the ketone.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Monosaccharides: L-Enantiomers

A

The highest-numbered hydroxyl group substituent is located on the left side of the projection.

The highest-numbered carbon stereocenter has the S-conformation.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Monosaccharides: D-Enantiomers

A

The highest-numbered hydroxyl group substituent is located on the right side of the projection.

The highest-numbered carbon stereocenter has the R-conformation.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Five-Membered Cyclic Hemiacetal

A

Furanose

The five-membered furanose ring is more stable than the six-membered pyranose ring.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Six-Membered Cyclic Hemiacetal

A

Pyranose

The six-membered pyranose ring is more stable than the five-membered furanose ring.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Stereoselectivity of Pyranose

A

The pyranose stereoisomer that places the C1 hydroxyl group at the equitorial position (i.e. the R-isomer) will be the dominant product.

The placement of the C1 hydroxyl group at the equitorial position minimizes steric effects.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Stereochemsitry of Cyclic Hemiacetal Formation

A

Hemiacetal cyclization forms a pair of diasteromers (i.e. anomers) due to stereoisomerization at the aldehydic (C1) carbon.

The new stereocenter of hemiacetal cyclization is formed at the former aldehydic carbon.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Anomer

A

A stereoisomer of a cyclic hemiacetal

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Mutarotation

A

The interconversion between the α-anomer and β-anomer of a cyclic hemiacetal.

The mutarotation interconversions occur while the aldoses are in the acyclic form. (The α-anomer and β-anomer of the cyclic aldose are in equilibrium in solution).

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

α-Anomer vs. β-Anomer

Cyclic Hemiacetals

A
  • α-Anomer: The C1 hydroxyl group and highest-numbered hydroxyl group are placed on the same side of the cyclic projection.
  • β-Anomer: The C1 hydroxyl group and highest-numbered hydroxyl group are placed on opposite sides of the cyclic projection.

The stereochemistry/configuration at the anomeric carbon is the ONLY difference between the α-anomer and β-anomer.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Anomeric Carbon

Cyclic Hemiacetals

A

C1 Carbon

The anomeric carbon is the hemiacetal carbon of a cyclic monosaccharide.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Stereoisomer Priority Trend

A

—OH > —COOH > —CH2OH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Acyclic Monosaccharide → Cyclic Hemiacetal

A

Hemiacetal Cyclization

The hemiacetal cyclization reaction can occur under basic conditions OR acidic conditions.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Cyclic Hemiacetal → Acyclic Monosaccharide

A

Cyclic Hemiacetal Ring-Opening

The hemiacetal cyclization reaction can is reversible.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Reagents: Hemiacetal Cyclization

A
  • Acid-Catalyzed: H+, H2O
  • Base-Catalyzed: OH–, H2O

The hemiacetal cyclization reaction CAN occur under neutral conditions (but the cyclization rate is much slower without a catalyst).

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Enantiomer-Switching

D-Saccharide → L-Saccharide

A

All chiral centers within the aldose/ketose are inverted to the opposite stereoisomer configuration.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Fischer Projections: Monosaccharides

A
  • Carbon atoms are numbered top-to-bottom.
  • Horizontal bonds are depicted as wedges.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Determining Anomer Stereochemistry

A
  • α-Anomer: The anomeric/hemiacetal carbon and the highest-numbered carbon have OPPOSITE stereoisomer configuration.
  • β-Anomer: The anomeric/hemiacetal carbon and the highest-numbered carbon have the SAME stereoisomer configuration.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Mutarotation: Interconversion Explanation

A

Interconversion results from the varying plane-of-attack of the alcohol/oxide group to form the cyclic hemiacetal. (Attack of the alcohol/oxide group from above the molecule plane OR below the molecule plane creates differing anomeric forms of the cyclic monosaccharide.)

Both cyclic anomeric forms of the monosaccharide are in equilibrium with the acyclic aldose conformation.

25
Q

Plane Analysis: Determining Anomer Stereochemistry

This plane analysis technique functions ONLY when the endocyclic Oxygen is bonded to the highest-numbered carbon.

A
  • α-Anomer: The anomeric hydroxyl group and the terminal CH2OH group are on OPPOSITE planes of the cyclic configuration.
  • β-Anomer: The anomeric hydroxyl group and the terminal CH2OH group are on the SAME plane of the cyclic configuration.
26
Q

Strength of Aldose Oxidants

A

Bromine-Water < Nitric Acid < Periodic Acid

  • Bromine-Water: Oxidation of Terminal Aldehyde ONLY.
  • Nitric Acid: Oxidation of Terminal Aldehyde AND Terminal Alcohol.
  • Periodic Acid: Oxidation of ALL Carbon-Carbon Bonds
27
Q

Aldose → Aldonic Acid

A

Bromine-Water Oxidation

Weak Oxidation

The Bromine-Water Oxidation reaction converts the terminal aldehyde group to a carboxylic acid group.

28
Q

Aldonic Acid

A

A sugar-acid monosaccharide containing one terminal carboxylic acid group.

29
Q

Aldose → Aldaric Acid

A

Nitric Acid Oxidation

Medium Oxidation

The Nitric Acid Oxidation reaction converts the terminal aldehyde group AND terminal alcohol group to carboxylic acid groups.

30
Q

Aldose → Carboxylic Acid + Aldehyde

A

Periodic Acid Oxidation

Strong Oxidation

The Periodic Acid Oxidation reaction oxidizes ALL carbon-carbon (C—C) bonds to C—OH or C=O bonds

31
Q

Ketose → Caboxylic Acid + Aldehyde + CO2

A

Periodic Acid Oxidation

Strong Oxidation

The Periodic Acid Oxidation reaction oxidizes ALL carbon-carbon (C—C) bonds to C—OH or C=O bonds

32
Q

Aldaric Acid

A

A sugar-acid monosaccharide containing two terminal carboxylic acid groups.

33
Q

Reagents: Bromine-Water Oxidation

A

Br2, H2O

34
Q

Reagents: Nitric Acid Oxidation

A

HNO3, H2O

35
Q

Reagents: Periodic Acid Oxidation

A

HIO4 (Excess)

36
Q

Product Identities: Periodic Acid Oxidation

A
  • Ketones → CO2
  • Aldehydes → Formic Acid
  • 2° Alcohols → Formic Acid
  • 1° Alcohols → Aldehydes
37
Q

Formic Acid vs. Formaldehyde

A
  • Formic Acid: 1-Carbon Carboxylic Acid
  • Formaldehyde: 1-Carbon Aldehyde
38
Q

Aldose → Alditol

A

NaBH4 Reduction

The NaBH4 Reduction reaction reduces the terminal aldehyde group to an alcohol group.

39
Q

Ketose → Alditol

A

NaBH4 Reduction

The NaBH4 Reduction reaction reduces the ketone group to a 2° alcohol group.

40
Q

Reagents: NaBH4 Reduction

A

NaBH4, CH3OH (Solvent)

41
Q

Mild/Weak Oxidants

A
  • Bromine-Water: Br2, H2O
  • Silver: Ag(I)
  • Copper: Cu(II)
42
Q

Why are 2° alcohol groups NOT oxidized during Nitric Acid Oxidation?

A

The 2° alcohol groups are too sterically hindered to interact with the nitric acid oxidant.

43
Q

Reaction: Sugar Chain Extension

A

Cyanohydrin Formation-Reduction

  • The Cyanohydrin Formation-Reduction reaction creates two new sugars (due to formation of a stereocenter) that are one carbon longer than the starting aldose.
44
Q

Reagents: Cyanohydrin Formation-Reduction

Starting Material: Aldose

A
  1. HCN
  2. H2, Pd
  3. H2O
45
Q

Mechanism: Cyanohydrin Formation-Reduction

A
  1. HCN Attack at Aldehydic Carbon
  2. Hydrogenation across C≡N Bond
  3. Hydration of Imine Group
46
Q

Pyranose → Multi-Pyranoside

Furanose → Multi-Furanoside

A

Base-Catalyzed William Ether Synthesis

The base-catalyzed William Ether Synthesis reaction converts ALL hydroxyl groups (on the pyranose) to methoxy groups.

47
Q

Pyranose → Mono-Pyranoside

Furanose → Mono-Furanoside

A

Acid-Catalyzed Anomeric Hydroxyl Methylation

The acid-catalyzed Anomeric Hydroxyl Methylation reaction converts ONLY the anomeric hydroxyl group to a methoxy group (i.e. it converts the hemiacetal group to an acetal group).

48
Q

Reagents: Base-Catalyzed William Ether Synthesis

Saccharide Acetal Formation

A

NaOH, (CH3)2SO4

49
Q

Reagents: Acid-Catalyzed Anomeric Hydroxyl Methylation

Saccharide Acetal Formation

A

CH3OH, H3O+

50
Q

Glycoside

A

Acetal-Containing Monosaccharide

51
Q

Stability of Glycoside Acetals

A
  • The acetal group is stable under basic conditions AND neutral conditions.
  • The acetal group is cleaved under acidic conditions.

  • Mutarotation of glycoside acetals does NOT occur under basic conditions or neutral conditions.
  • Mutarotation of glycoside acetals DOES occur under acidic conditions (after the acetal is converted to a hemiacetal).
52
Q

Reagents: Cleavage of Glycoside Acetal

A

H3O+, Δ (Heat)

53
Q

Stability of Cyclic Acetals

A

Cyclic acetals are stable under basic conditions AND neutral conditions.

54
Q

1,2-Diol → Cyclic Acetal

A

Ketone, H3O+

55
Q

Methods of Sugar Hydoxyl Protection

A
  • Cyclic Acetal Synthesis via 1,2-Diol
  • Base-Catalyzed William Ether Synthesis
  • Acid-Catalyzed Anomeric Hydroxyl Methylation
56
Q

Disaccharide → Monosaccharide

A

Ether Linkage Cleavage

The α-anomer AND β-anomer of the monopyranose are formed. (The anomer with the anomeric hydroxyl group placed at the equitorial position will predominate.)

57
Q

Reagents: Ether Linkage Cleavage

A

H3O+

58
Q

Why are hemiacetal —OH groups more reactive than simple —OH groups under acidic conditions?

Pyranose Cleavage

A

Reaction of the hemiacetal —OH group with acid (i.e. protonation) results in a more stable oxocarbenium intermediate form.

Reaction of the simple —OH group with acid results in a less stable carbocation intermediate form.

Organic Chemistry 2 (12 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions